HNO2(aq) ⇄ H+(aq) + NO2−(aq) Ka = 4.0 × 10−4 On the basis of the information above, what is the appr — Acids and Bases Chemistry Question

STEPS:

1. Identify the balanced dissociation equation and the given values:
We are given the equilibrium equation for the ionization of nitrous acid ($\text{HNO}_2$) in water:
$$\text{HNO}_2(aq) \rightleftharpoons \text{H}^+(aq) + \text{NO}_2^-(aq) \quad K_a = 4.0 \times 10^{-4}$$
The initial concentration of nitrous acid is $[\text{HNO}_2]_{\text{initial}} = 1.0\text{ M}$.

2. Set up the acid dissociation constant ($K_a$) expression:
The equilibrium expression for this weak acid ionization is:
$$K_a = \frac{[\text{H}^+][\text{NO}_2^-]}{[\text{HNO}_2]}$$

3. Define equilibrium concentrations using an ICE table approach:
* Let $x$ represent the concentration of $\text{HNO}_2$ that ionizes to reach equilibrium:
* $[\text{H}^+]_{\text{equilibrium}} = x$
* $[\text{NO}_2^-]_{\text{equilibrium}} = x$
* $[\text{HNO}_2]_{\text{equilibrium}} = 1.0 - x$

4. Substitute the equilibrium values into the $K_a$ expression:
$$4.0 \times 10^{-4} = \frac{x^2}{1.0 - x}$$

5. Apply the weak acid approximation:
Since the equilibrium constant is small ($K_a = 4.0 \times 10^{-4} \ll 1$), we can assume that the amount of acid that ionizes ($x$) is negligible compared to the initial concentration of $1.0\text{ M}$ ($1.0 - x \approx 1.0$):
$$4.0 \times 10^{-4} \approx \frac{x^2}{1.0}$$

6. Solve for $x$ (which equals the equilibrium $[\text{H}^+]$ concentration):
$$x^2 \approx 4.0 \times 10^{-4}$$
$$x = \sqrt{4.0 \times 10^{-4}} = \mathbf{2.0 \times 10^{-2}\text{ M}} \quad (\text{or } 0.020\text{ M})$$
*(Note: Because $0.020\text{ M}$ is only $2\%$ of the initial concentration, this approximation is highly valid and easily passes the standard $5\%$ rule).*

7. Calculate the percent ionization:
The percent ionization is the ratio of the concentration of the ionized acid at equilibrium to the initial concentration of the acid, expressed as a percentage:
$$\text{Percent Ionization} = \frac{[\text{H}^+]_{\text{equilibrium}}}{[\text{HNO}_2]_{\text{initial}}} \times 100\%$$
$$\text{Percent Ionization} = \frac{0.020\text{ M}}{1.0\text{ M}} \times 100\% = \mathbf{2.0\%}$$
This identifies Option D as the correct answer.

*

WHY_OTHERS_WRONG:

• Option A is incorrect (0.00040%): A student might choose this value by directly multiplying the $K_a$ value ($4.0 \times 10^{-4}$) by $100\%$ without setting up the equilibrium calculation first, and then making a decimal error.
• Option B is incorrect (0.020%): This represents the calculated molar concentration of $\text{H}^+$ at equilibrium ($0.020\text{ M}$). A student would choose this if they solved for $x$ correctly but forgot to multiply by $100\%$ to convert the decimal fraction into a percentage.
• Option C is incorrect (0.040%): A student might arrive at this value by mistakenly dividing the $K_a$ value ($4.0 \times 10^{-4}$) by the initial concentration ($1.0\text{ M}$) and then multiplying by $100\%$: $\frac{4.0 \times 10^{-4}}{1.0} \times 100\% = 0.040\%$. This error skips the step of relating the equilibrium concentrations of the products as $x^2$.